题目内容
计算:(1)(| x-2 |
| x+2 |
| x+2 |
| x-2 |
| x2-2x |
| x2 |
(2)(
| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| x-4 |
| x |
分析:(1)先把括号内的分式通分合并,然后把各分式的分子分母因式分解,然后约去即可.
(2)先把各分式的分子分母因式分解,找到括号内的最简公分母,通分合并,再约分即可.
(2)先把各分式的分子分母因式分解,找到括号内的最简公分母,通分合并,再约分即可.
解答:解:(1)原式=
•
=
•
=-
;
(2)原式=[
-
]•
=
•
=
.
| (x-2) 2-(x+2)2 |
| (x+2)(x-2) |
| x(x-2) |
| x2 |
=
| -8x |
| (x+2)(x-2) |
| x(x-2) |
| x2 |
=-
| 8 |
| x+2 |
(2)原式=[
| x+2 |
| x(x-2) |
| x-1 |
| (x-2)2 |
| x |
| x-4 |
=
| x-4 |
| x(x-2) 2 |
| x |
| x-4 |
=
| 1 |
| x2-4x+4 |
点评:本题考查了分式的混合运算:先把括号内的分式通分合并,再把各分式的分子分母因式分解,然后约去即可.
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